Which determinant could we find?

Question

$A$ and $B$ are matrices and I found the determinants of

$$A + B,\, A - B,\, AB,\, A^{-1},\, B^T.$$

If we know the determinants of $A$ and $B$ but don't remember the matrices $A$ and $B$, which of the determinants above could we have found?

I think it's only $B^T$ and $A^{-1}$. Can someone tell me if that is correct?

Also, $;A;$ is regular (invertible, non-singular) iff $;\det A\neq 0\implies \det A^{-1}=\left(\det A \right)^{-1};$ . Also, it is always true that $;\det A=\det A^t;$ — Timbuc, Feb 20 '15 at 09:32
See also: http://math.stackexchange.com/questions/466043/deta-b-deta-detb — Martin Sleziak, Feb 20 '15 at 11:30

score 1 · Answer 1 · answered Feb 20 '15 at 10:45

From some basic facts about determinants we know that:

$|B|=|B^T|$ (ProofWiki, Google)
$|AB|=|A|\cdot|B|$ (ProofWiki, Google)
$|A^{-1}|=|A|^{-1}$ (ProofWiki, Google)

So if we know $|A|$ and $|B|$, we also know the determinants of $AB$, $B^T$ and $A^{-1}$.

To show that we cannot say anything about the value of $|A\pm B|$ from the values of $|A|$ and $|B|$, we can simply try some examples:

For $A=B=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ we have $|A|=|B|=1$ and $|A+B|=4$ and $|A-B|=0$.
For $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $B=\begin{pmatrix}1&0\\1&1\end{pmatrix}$ we again have $|A|=|B|=1$. But in this case $|A+B|=\begin{vmatrix}2&1\\1&2\end{vmatrix}=3$ and $|A-B|=\begin{vmatrix}0&1\\-1&0\end{vmatrix}=1$.

So in both cases $|A|=|B|=1$, but determinants of $A\pm B$ have different values. Which means that they are not determined by the values of $|A|$ and $|B|$.

Which determinant could we find?

1 Answers1